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        <hr>
<p>2017.08.17更新</p>
<hr>
<h1 id="背包问题"><a href="#背包问题" class="headerlink" title="背包问题"></a>背包问题</h1><h2 id="01背包"><a href="#01背包" class="headerlink" title="01背包"></a>01背包</h2><h3 id="状态"><a href="#状态" class="headerlink" title="状态"></a>状态</h3><p><code>dp[i][j]</code>表示前i个物品装到剩余容量为j时的最大价值</p>
<h3 id="状态转移方程"><a href="#状态转移方程" class="headerlink" title="状态转移方程"></a>状态转移方程</h3><p><code>dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weight[i - 1]] + value[i - 1])</code><br>第<code>i</code>个物品装或者不装入背包，不装价值为<code>dp[i-1][j]</code>,装表示剩下<code>i-1</code>个物品装入<code>j-weight[i-1]</code>重的最大价值，加上<code>value[i-1]</code><br>注意讨论前<code>i</code>个物品装入背包的时候， 其实是在考查第<code>i-1</code>个物品装不装入背包（因为物品是从0开始编号的）<br><a id="more"></a></p>
<h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">public</span> <span class="class"><span class="keyword">class</span> <span class="title">Pack_01</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">	<span class="function"><span class="keyword">public</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title">main</span><span class="params">(String[] args)</span> </span>&#123;</span><br><span class="line">		<span class="comment">// TODO Auto-generated method stub</span></span><br><span class="line"><span class="comment">//		int n = 5;// 物品个数</span></span><br><span class="line"><span class="comment">//		int[] value = &#123; 48, 40, 12, 8, 7 &#125;;// 物品价值</span></span><br><span class="line"><span class="comment">//		int[] weight = &#123; 6, 5, 2, 1, 1 &#125;;// 物品重量</span></span><br><span class="line"><span class="comment">//		int capacity = 8;// 背包容量</span></span><br><span class="line"></span><br><span class="line">		<span class="keyword">int</span> n = <span class="number">4</span>;</span><br><span class="line">		<span class="keyword">int</span>[] value = &#123; <span class="number">10</span>, <span class="number">40</span>, <span class="number">30</span>, <span class="number">50</span> &#125;;</span><br><span class="line">		<span class="keyword">int</span>[] weight = &#123; <span class="number">5</span>, <span class="number">4</span>, <span class="number">6</span>, <span class="number">3</span> &#125;;</span><br><span class="line">		<span class="keyword">int</span> capacity = <span class="number">12</span>;</span><br><span class="line">		<span class="keyword">int</span> res = pack_01(n, capacity, weight, value);</span><br><span class="line">		System.out.println(res);</span><br><span class="line"></span><br><span class="line">	&#125;</span><br><span class="line"></span><br><span class="line">	<span class="function"><span class="keyword">public</span> <span class="keyword">static</span> <span class="keyword">int</span> <span class="title">pack_01</span><span class="params">(<span class="keyword">int</span> n, <span class="keyword">int</span> capacity, <span class="keyword">int</span>[] weight, <span class="keyword">int</span>[] value)</span> </span>&#123;</span><br><span class="line">		<span class="keyword">int</span>[][] dp = <span class="keyword">new</span> <span class="keyword">int</span>[n + <span class="number">1</span>][capacity + <span class="number">1</span>];</span><br><span class="line">		<span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">1</span>; i &lt; dp.length; i++) &#123;<span class="comment">//</span></span><br><span class="line">			<span class="keyword">for</span> (<span class="keyword">int</span> j = <span class="number">1</span>; j &lt; dp[<span class="number">0</span>].length; j++) &#123;</span><br><span class="line">				<span class="keyword">if</span> (j &gt;= weight[i - <span class="number">1</span>]) &#123;</span><br><span class="line">					dp[i][j] = Math.max(dp[i - <span class="number">1</span>][j], dp[i - <span class="number">1</span>][j - weight[i - <span class="number">1</span>]] + value[i - <span class="number">1</span>]);</span><br><span class="line">				&#125; <span class="keyword">else</span> &#123;</span><br><span class="line">					dp[i][j] = dp[i - <span class="number">1</span>][j];<span class="comment">// i只与i-1有关</span></span><br><span class="line">				&#125;</span><br><span class="line">			&#125;</span><br><span class="line">		&#125;</span><br><span class="line">		<span class="keyword">return</span> dp[n][capacity];</span><br><span class="line">	&#125;</span><br><span class="line"></span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<h2 id="完全背包"><a href="#完全背包" class="headerlink" title="完全背包"></a>完全背包</h2><h3 id="状态-1"><a href="#状态-1" class="headerlink" title="状态"></a>状态</h3><p><code>dp[i][j]</code>表示前i个物品装到剩余容量为j时的最大价值</p>
<h3 id="状态转移方程-1"><a href="#状态转移方程-1" class="headerlink" title="状态转移方程"></a>状态转移方程</h3><p><code>dp[i][j] = max(dp[i - 1][j - num * weight[i - 1]] + num * value[i - 1]) (0&lt;=num * weight[i - 1]&lt;=j)</code></p>
<h3 id="代码-1"><a href="#代码-1" class="headerlink" title="代码"></a>代码</h3><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">public</span> <span class="class"><span class="keyword">class</span> <span class="title">Pack_full</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">	<span class="function"><span class="keyword">public</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title">main</span><span class="params">(String[] args)</span> </span>&#123;</span><br><span class="line">		<span class="comment">// TODO Auto-generated method stub</span></span><br><span class="line">		<span class="keyword">int</span> n = <span class="number">5</span>;<span class="comment">// 物品个数</span></span><br><span class="line">		<span class="keyword">int</span>[] value = &#123; <span class="number">48</span>, <span class="number">40</span>, <span class="number">12</span>, <span class="number">8</span>, <span class="number">7</span> &#125;;<span class="comment">// 物品价值</span></span><br><span class="line">		<span class="keyword">int</span>[] weight = &#123; <span class="number">6</span>, <span class="number">5</span>, <span class="number">2</span>, <span class="number">1</span>, <span class="number">1</span> &#125;;<span class="comment">// 物品重量</span></span><br><span class="line">		<span class="keyword">int</span> capacity = <span class="number">8</span>;<span class="comment">// 背包容量</span></span><br><span class="line"></span><br><span class="line">		<span class="comment">// int n = 4;</span></span><br><span class="line">		<span class="comment">// int[] value = &#123; 10, 5, 30, 40 &#125;;</span></span><br><span class="line">		<span class="comment">// int[] weight = &#123; 5, 4, 6, 3 &#125;;</span></span><br><span class="line">		<span class="comment">// int capacity = 12;</span></span><br><span class="line">		<span class="keyword">int</span> res = pack_full(n, capacity, weight, value);</span><br><span class="line">		System.out.println(res);</span><br><span class="line">	&#125;</span><br><span class="line"></span><br><span class="line">	<span class="function"><span class="keyword">public</span> <span class="keyword">static</span> <span class="keyword">int</span> <span class="title">pack_full</span><span class="params">(<span class="keyword">int</span> n, <span class="keyword">int</span> capacity, <span class="keyword">int</span>[] weight, <span class="keyword">int</span>[] value)</span> </span>&#123;</span><br><span class="line">		<span class="keyword">int</span>[][] dp = <span class="keyword">new</span> <span class="keyword">int</span>[n + <span class="number">1</span>][capacity + <span class="number">1</span>];</span><br><span class="line">		<span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">1</span>; i &lt; dp.length; i++) &#123;</span><br><span class="line">			<span class="keyword">for</span> (<span class="keyword">int</span> j = <span class="number">1</span>; j &lt; dp[<span class="number">0</span>].length; j++) &#123;</span><br><span class="line">				<span class="keyword">for</span> (<span class="keyword">int</span> num = <span class="number">0</span>; j &gt;= num * weight[i - <span class="number">1</span>]; num++) &#123;</span><br><span class="line">					dp[i][j] = Math.max(dp[i][j], dp[i - <span class="number">1</span>][j - num * weight[i - <span class="number">1</span>]] + num * value[i - <span class="number">1</span>]);</span><br><span class="line">				&#125;</span><br><span class="line">			&#125;</span><br><span class="line">		&#125;</span><br><span class="line">		<span class="keyword">return</span> dp[n][capacity];</span><br><span class="line">	&#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<hr>
<p>2017.08.18更新</p>
<hr>
<h2 id="硬币找零-方案数"><a href="#硬币找零-方案数" class="headerlink" title="硬币找零(方案数)"></a>硬币找零(方案数)</h2><h3 id="状态-2"><a href="#状态-2" class="headerlink" title="状态"></a>状态</h3><p><code>dp[i][j</code>]表示使用<code>changes[0-i]</code>硬币兑换<code>j</code>元的方法总数</p>
<h3 id="分析"><a href="#分析" class="headerlink" title="分析"></a>分析</h3><p>使用i=0的钱币兑换，只有changes[0]的整数倍的金额才能有1种方法<br><code>dp[0][j * changes[0]] = 1</code></p>
<h3 id="状态转移方程-2"><a href="#状态转移方程-2" class="headerlink" title="状态转移方程"></a>状态转移方程</h3><p>不装入第i种钱币，即使用0~i-1种钱币组成j的方法数；装入i钱币，使用0~i的钱币组成j-changes[i]金额的方法数<br><code>dp[i][j] = dp[i - 1][j] +  dp[i][j - changes[i]]</code></p>
<h3 id="代码-2"><a href="#代码-2" class="headerlink" title="代码"></a>代码</h3><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">//链接：https://www.nowcoder.com/questionTerminal/185dc37412de446bbfff6bd21e4356ec</span></span><br><span class="line"><span class="comment">//来源：牛客网</span></span><br><span class="line"><span class="comment">//</span></span><br><span class="line"><span class="comment">//有一个数组changes，changes中所有的值都为正数且不重复。每个值代表一种面值的货币，每种面值的货币可以使用任意张，对于一个给定值x，请设计一个高效算法，计算组成这个值的方案数。</span></span><br><span class="line"><span class="comment">//给定一个int数组changes，代表所以零钱，同时给定它的大小n，另外给定一个正整数x，请返回组成x的方案数，保证n小于等于100且x小于等于10000。</span></span><br><span class="line"><span class="comment">//测试样例：</span></span><br><span class="line"><span class="comment">//[5,10,25,1],4,15</span></span><br><span class="line"><span class="comment">//返回：</span></span><br><span class="line"><span class="comment">//6</span></span><br><span class="line"></span><br><span class="line"><span class="keyword">public</span> <span class="class"><span class="keyword">class</span> <span class="title">ChangeMoney</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">	<span class="function"><span class="keyword">public</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title">main</span><span class="params">(String[] args)</span> </span>&#123;</span><br><span class="line">		<span class="comment">// TODO Auto-generated method stub</span></span><br><span class="line">		<span class="keyword">int</span>[] changes = &#123; <span class="number">5</span>, <span class="number">10</span>, <span class="number">25</span>, <span class="number">1</span> &#125;;</span><br><span class="line">		<span class="keyword">int</span> n = <span class="number">4</span>;</span><br><span class="line">		<span class="keyword">int</span> money = <span class="number">15</span>;</span><br><span class="line">		<span class="keyword">int</span> res = change(changes, n, money);</span><br><span class="line">		System.out.println(res);</span><br><span class="line"></span><br><span class="line">	&#125;</span><br><span class="line"></span><br><span class="line">	<span class="function"><span class="keyword">public</span> <span class="keyword">static</span> <span class="keyword">int</span> <span class="title">change</span><span class="params">(<span class="keyword">int</span>[] changes, <span class="keyword">int</span> n, <span class="keyword">int</span> money)</span> </span>&#123;</span><br><span class="line">		<span class="keyword">int</span>[][] dp = <span class="keyword">new</span> <span class="keyword">int</span>[n][money + <span class="number">1</span>];<span class="comment">// dp[i][j]表示使用changes[0-i]硬币兑换j元的方法总数。</span></span><br><span class="line">		<span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; n; i++) &#123; <span class="comment">// j=0表示钱为0，组成0元的方法数为1</span></span><br><span class="line">			dp[i][<span class="number">0</span>] = <span class="number">1</span>;</span><br><span class="line">		&#125;</span><br><span class="line">		<span class="keyword">for</span> (<span class="keyword">int</span> j = <span class="number">0</span>; j * changes[<span class="number">0</span>] &lt; money + <span class="number">1</span>; j++) &#123;<span class="comment">// 使用i=0的钱币兑换，只有changes[0]的整数倍的金额才能有1种方法</span></span><br><span class="line">			dp[<span class="number">0</span>][j * changes[<span class="number">0</span>]] = <span class="number">1</span>;</span><br><span class="line">		&#125;</span><br><span class="line">		<span class="comment">// 填表</span></span><br><span class="line">		<span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">1</span>; i &lt; dp.length; i++) &#123;<span class="comment">// 数组长度1~n-1在changes数组中有效</span></span><br><span class="line">			<span class="keyword">for</span> (<span class="keyword">int</span> j = <span class="number">1</span>; j &lt; dp[<span class="number">0</span>].length; j++) &#123;</span><br><span class="line">				<span class="comment">// 不装入第i种钱币，即使用0~i-1种钱币组成j的方法数；装入i钱币，使用0~i的钱币组成j-changes[i]金额的方法数</span></span><br><span class="line">				dp[i][j] = dp[i - <span class="number">1</span>][j] + (j - changes[i] &gt;= <span class="number">0</span> ? dp[i][j - changes[i]] : <span class="number">0</span>);</span><br><span class="line">			&#125;</span><br><span class="line">		&#125;</span><br><span class="line">		<span class="keyword">return</span> dp[n - <span class="number">1</span>][money];</span><br><span class="line">	&#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<h2 id="硬币找零-最少硬币数"><a href="#硬币找零-最少硬币数" class="headerlink" title="硬币找零(最少硬币数)"></a>硬币找零(最少硬币数)</h2><h2 id="多重背包"><a href="#多重背包" class="headerlink" title="多重背包"></a>多重背包</h2><hr>
<p>2017-08-28更新</p>
<hr>
<h2 id="最长公共子序列"><a href="#最长公共子序列" class="headerlink" title="最长公共子序列"></a>最长公共子序列</h2><h2 id="最长递增子序列"><a href="#最长递增子序列" class="headerlink" title="最长递增子序列"></a>最长递增子序列</h2><h3 id="动态规划法"><a href="#动态规划法" class="headerlink" title="动态规划法"></a>动态规划法</h3><figure class="highlight javascript"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">//普通dp算法，复杂度为n^2</span></span><br><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">longest_increasing_subsequence_1</span>(<span class="params">arr</span>) </span>&#123;</span><br><span class="line">    <span class="comment">//时间复杂度为n^2</span></span><br><span class="line">    <span class="keyword">let</span> dp = [];</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">0</span>; i &lt; arr.length; i++) &#123;</span><br><span class="line">        dp[i] = <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">let</span> max = <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">0</span>; i &lt; arr.length; i++) &#123;</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">let</span> j = <span class="number">0</span>; j &lt; i; j++) &#123;</span><br><span class="line">            <span class="keyword">if</span> (arr[j] &lt; arr[i]) &#123;</span><br><span class="line">                dp[i] = <span class="built_in">Math</span>.max(dp[j] + <span class="number">1</span>, dp[i]);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        max = <span class="built_in">Math</span>.max(max, dp[i]);</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> max;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<h3 id="改进的二分查找法"><a href="#改进的二分查找法" class="headerlink" title="改进的二分查找法"></a>改进的二分查找法</h3><figure class="highlight javascript"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">//二分查找，复杂度为nlogn</span></span><br><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">longest_increasing_subsequence_2</span>(<span class="params">arr</span>) </span>&#123;</span><br><span class="line">    <span class="comment">//时间复杂度为nlogn</span></span><br><span class="line">	<span class="comment">//minNum存储的是长度为 index+1 的最小末尾值</span></span><br><span class="line">	<span class="comment">//当arr[i]比minNum数组的最后一个数还大时</span></span><br><span class="line">	<span class="comment">//minNum数组长度加1,其最小末尾为arr[i]</span></span><br><span class="line">	<span class="comment">//这样保证minNum数组的值是有序排列的</span></span><br><span class="line">	<span class="comment">//当arr[i]比比minNum数组的最后一个数小时,</span></span><br><span class="line">	<span class="comment">//对minNum有序数组进行二分查找</span></span><br><span class="line">	<span class="comment">//并将最终找到的右边界返回，将其值改为arr[i]</span></span><br><span class="line">	<span class="comment">//保证了minNum数组存储的一直为最小末尾</span></span><br><span class="line">	<span class="comment">//即相同长度的子序列，选择末尾较小的填入</span></span><br><span class="line">	<span class="comment">//以保证minNum数组一直有序</span></span><br><span class="line">    <span class="keyword">let</span> minNum = [];</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">0</span>; i &lt; arr.length; i++) &#123;</span><br><span class="line">        length = minNum.length;</span><br><span class="line">        <span class="keyword">if</span> (arr[i] &gt; minNum[length - <span class="number">1</span>] || minNum[length - <span class="number">1</span>] == <span class="literal">undefined</span>) &#123;</span><br><span class="line">            minNum[length] = arr[i];</span><br><span class="line">        &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">            <span class="comment">//二分查找</span></span><br><span class="line">            <span class="keyword">let</span> change_index = binary_search(minNum,arr[i]);</span><br><span class="line">            minNum[change_index] = arr[i];</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> minNum.length;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">binary_search</span>(<span class="params">arr,search_num</span>)</span>&#123;</span><br><span class="line">    <span class="keyword">let</span> seek_left = <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">let</span> seek_right = arr.length - <span class="number">1</span>;</span><br><span class="line">    <span class="keyword">let</span> half_length = <span class="built_in">Math</span>.floor((seek_left + seek_right) / <span class="number">2</span>);</span><br><span class="line">    <span class="keyword">while</span> ((seek_right - seek_left) &gt; <span class="number">1</span>) &#123;</span><br><span class="line">        <span class="keyword">if</span> (arr[half_length] &lt; search_num) &#123;</span><br><span class="line">            seek_left = <span class="built_in">Math</span>.floor((seek_left + seek_right) / <span class="number">2</span>);</span><br><span class="line">        &#125; <span class="keyword">else</span> <span class="keyword">if</span> (arr[half_length] &gt; search_num) &#123;</span><br><span class="line">            seek_right = <span class="built_in">Math</span>.floor((seek_left + seek_right) / <span class="number">2</span>);</span><br><span class="line">        &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">            seek_right = half_length;</span><br><span class="line">            <span class="keyword">break</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        half_length = <span class="built_in">Math</span>.floor((seek_left + seek_right) / <span class="number">2</span>);</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> seek_right;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<h2 id="地牢游戏"><a href="#地牢游戏" class="headerlink" title="地牢游戏"></a>地牢游戏</h2><figure class="highlight javascript"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">function</span> <span class="title">count</span>(<span class="params">arr</span>) </span>&#123;</span><br><span class="line"></span><br><span class="line">    <span class="comment">//dp[i][j]表示可以从点(i,j)到终点所需的最小能量</span></span><br><span class="line">    <span class="keyword">var</span> dp = [];</span><br><span class="line">    <span class="comment">//初始化数组中的每个值均为0</span></span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">var</span> i = <span class="number">0</span>; i &lt; arr.length; i++) &#123;</span><br><span class="line">        dp[i] = [];</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">var</span> j = <span class="number">0</span>; j &lt; arr[<span class="number">0</span>].length; j++) &#123;</span><br><span class="line">            dp[i][j] = <span class="number">0</span>;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="comment">//从后往前</span></span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">var</span> i = arr.length - <span class="number">1</span>; i &gt;= <span class="number">0</span>; i--) &#123;</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">var</span> j = arr[<span class="number">0</span>].length - <span class="number">1</span>; j &gt;= <span class="number">0</span>; j--) &#123;</span><br><span class="line">            <span class="comment">//1.边缘情况和初始值</span></span><br><span class="line">            <span class="keyword">if</span> (i == arr.length - <span class="number">1</span> &amp;&amp; j == arr[<span class="number">0</span>].length - <span class="number">1</span>) &#123;</span><br><span class="line">                dp[i][j] = <span class="built_in">Math</span>.max(<span class="number">1</span>, <span class="number">1</span> - arr[i][j]);</span><br><span class="line">            &#125; <span class="keyword">else</span> <span class="keyword">if</span> (i == arr.length - <span class="number">1</span> &amp;&amp; j != arr[<span class="number">0</span>].length - <span class="number">1</span>) &#123;</span><br><span class="line">                dp[i][j] = <span class="built_in">Math</span>.max(<span class="number">1</span>, dp[i][j + <span class="number">1</span>] - arr[i][j]);</span><br><span class="line">            &#125; <span class="keyword">else</span> <span class="keyword">if</span> (i != arr.length - <span class="number">1</span> &amp;&amp; j == arr[<span class="number">0</span>].length - <span class="number">1</span>) &#123;</span><br><span class="line">                dp[i][j] = <span class="built_in">Math</span>.max(<span class="number">1</span>, dp[i + <span class="number">1</span>][j] - arr[i][j]);</span><br><span class="line">            &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">            <span class="comment">//2.正常情况</span></span><br><span class="line">            <span class="comment">//这个能量永远大于等于1，小于等于表示人已死</span></span><br><span class="line">            <span class="comment">//从后往前表示往后的路径已知，那么到达该点的最小能量值也可确定</span></span><br><span class="line">            <span class="comment">//从前往后因为无法确定之后的情况和路径所以不能判定该点的状态</span></span><br><span class="line">            <span class="comment">//表示从右下两个的位置所需的最小能量减去该点获得的能量的最小值</span></span><br><span class="line">            <span class="comment">//但需要比0大，有小于等于0的表示人已死</span></span><br><span class="line">            <span class="comment">//状态转移方程如下</span></span><br><span class="line">                dp[i][j] = <span class="built_in">Math</span>.max(<span class="number">1</span>, <span class="built_in">Math</span>.min(dp[i + <span class="number">1</span>][j], dp[i][j + <span class="number">1</span>]) - arr[i][j]);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> dp[<span class="number">0</span>][<span class="number">0</span>];</span><br><span class="line"></span><br><span class="line"></span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<h2 id="参考文献"><a href="#参考文献" class="headerlink" title="参考文献"></a>参考文献</h2><ol>
<li><a href="http://love-oriented.com/pack/" target="_blank" rel="noopener">背包问题九讲</a></li>
<li><a href="http://www.hawstein.com/posts/dp-knapsack.html" target="_blank" rel="noopener">动态规划之背包问题</a></li>
</ol>

      
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    <strong>本文作者：</strong>
    zxlg
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    <strong>本文链接：</strong>
    <a href="http://happylg.cn/2017/08/13/dynamic-programming/" title="动态规划">http://happylg.cn/2017/08/13/dynamic-programming/</a>
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            counter.save(null, {
              success: function(counter) {
                var $element = $(document.getElementById(url));
                $element.find('.leancloud-visitors-count').text(counter.get('time'));
              },
              error: function(counter, error) {
                console.log('Failed to save Visitor num, with error message: ' + error.message);
              }
            });
          } else {
            var newcounter = new Counter();
            /* Set ACL */
            var acl = new AV.ACL();
            acl.setPublicReadAccess(true);
            acl.setPublicWriteAccess(true);
            newcounter.setACL(acl);
            /* End Set ACL */
            newcounter.set("title", title);
            newcounter.set("url", url);
            newcounter.set("time", 1);
            newcounter.save(null, {
              success: function(newcounter) {
                var $element = $(document.getElementById(url));
                $element.find('.leancloud-visitors-count').text(newcounter.get('time'));
              },
              error: function(newcounter, error) {
                console.log('Failed to create');
              }
            });
          }
        },
        error: function(error) {
          console.log('Error:' + error.code + " " + error.message);
        }
      });
    }

    $(function() {
      var Counter = AV.Object.extend("Counter");
      if ($('.leancloud_visitors').length == 1) {
        addCount(Counter);
      } else if ($('.post-title-link').length > 1) {
        showTime(Counter);
      }
    });
  </script>



  

  

  

  

  

</body>
</html>
